/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Differentiate each function. \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Differentiate each function. \(f(x)=\frac{x^{-1}}{x+x^{-1}}\)

Short Answer

Expert verified
Differentiate using the quotient rule: \( f'(x) = \frac{0 \cdot (x^2 + 1) - 1 \cdot 2x}{(x^2 + 1)^2} = \frac{-2x}{(x^2 + 1)^2} \).

Step by step solution

01

Rewrite the Function

Before differentiating, let's rewrite the function to make it easier to handle fractions:\[ f(x) = \frac{x^{-1}}{x + x^{-1}} = \frac{1/x}{x + 1/x}. \]This is the same as simplifying the denominator:\[ f(x) = \frac{1/x}{\frac{x^2+1}{x}} = \frac{1/x \cdot x}{x^2 + 1} = \frac{1}{x^2 + 1}. \]
02

Differentiate the Inner Function

Differentiate the simplified version of the function using the derivative of a quotient. Apply the quotient rule: If \( h(x) = \frac{u(x)}{v(x)} \), then \( h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).Here, \( u(x) = 1 \) and \( v(x) = x^2 + 1 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
Differentiating functions that are in the form of one function divided by another can be tricky. This is where the Quotient Rule comes in. The Quotient Rule is a method for finding the derivative of a function that is the division of two differentiable functions. If we have a function \( h(x) = \frac{u(x)}{v(x)} \), then the derivative \( h'(x) \) is given by the formula:
  • \( h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)

This formula might look complicated, but it’s just following a systematic approach:
  • First, differentiate \( u(x) \), which is the numerator function.
  • Then, differentiate \( v(x) \), which is the denominator function.
  • Finally, plug these into the formula to find \( h'(x) \).
It’s important to remember that the order in which you subtract \( u(x)v'(x) \) from \( u'(x)v(x) \) matters. This is crucial for getting the correct derivative.
Function Simplification
Before applying any rules to differentiate, it's often a good idea to simplify the function if possible. Function Simplification involves rewriting the function to make differentiation easier. This could mean factoring out common terms, combining like terms, or rewriting fractions.
In our example, \( f(x)=\frac{x^{-1}}{x+x^{-1}} \) can be rewritten to make it more approachable. Rewriting it as \( \frac{1/x}{x + 1/x} \) for example, helps us see how the terms interact.
The expression in the denominator \( x + \frac{1}{x} \) can then be combined into a single fraction. Doing so yields \( \frac{x^2+1}{x} \). From there, multiplying the entire expression by \( x \) simplifies it to \( \frac{1}{x^2 + 1} \).
  • This simplification means that when we differentiate, we're dealing only with a simpler denominator, making the overall process more straightforward.
Remember: simplification doesn’t change the function itself, but it makes the process of finding derivatives much neater.
Mathematical Problem Solving
Solving mathematical problems, especially those involving calculus, involves several key steps and skills. Here’s a breakdown of the approach:
  • **Understand the problem:** Carefully read and interpret what you need to find. This exercise required differentiating a given function.
  • **Simplify where possible:** As we saw, simplifying the function \( f(x) = \frac{x^{-1}}{x+x^{-1}} \) was an important step to decrease complexity.
  • **Apply the right mathematical rules:** Use calculus rules effectively. In this case, the Quotient Rule is applicable since there's a division of two functions.
  • **Check your work:** Always revisit your computations. Errors can occur in simplification or differentiation, so a second look is always a smart move.

By practicing these steps, solving even the most complex calculus problems can become more manageable. It’s like solving a puzzle where each piece fits together seamlessly to reveal the bigger picture. Ensuring that every step is clear and executed accurately is key to mastering mathematical problem solving.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At the beginning of a trip, the odometer on a car reads \(30,680,\) and the car has a full tank of gas. At the end of the trip, the odometer reads \(31,077 .\) It takes 13.5 gal of gas to refill the tank. a) What is the average rate at which the car was traveling, in miles per gallon? b) What is the average rate of gas consumption in gallons per mile?

For each function, find the points on the graph at which the tangent line has slope 1 . $$ y=-0.01 x^{2}+2 x $$

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'Hôpital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 5}\left(\frac{x^{2}-25}{2 x-10}\right) $$

Suppose Fast Trends determines that the revenue, in dollars, from the sale of \(x\) iPod holders is given by $$ R(x)=-0.001 x^{2}+150 x $$ Find \(\frac{R(305)-R(300)}{305-300},\) and interpret the significance of this result to the company.

Find the interval(s) for which \(f^{\prime}(x)\) is positive. Use the derivative to help explain why \(k(x)=\frac{1}{x^{2}}\) decreases for all \(x\) in \((0, \infty)\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.